Math, asked by amitb52, 4 months ago

a circular disc of radious' s= 3cm has been
rimoved from a large disc of Radious R=10cm.
The centre of the hole so made it at
a=5cm. distance from the centre of original
disc. if the mass of the remaining disc is 182
gm. then calculate its m.I about an axis passing through
two centres . what is its M.I. about an axis
Passing thorough 'o'
and perpendicular to its plane​

Answers

Answered by SamridhiNainwal
0

Answer:

B

Step-by-step explanation:

Solution :

In Fig.

is the centre of circular disc of radius

and mass

is centre of disc of radius

, which is removed. If

is mass per unit area of disc, then

<br> Mass of disc removed,

<br> Mass of remaining disc,

<br>

<br> Let centre of mass of remaining disc be at

where

<br> As

<br>

<br>

Answered by pratham270378
0

Step-by-step explanation:

11th

Physics

Systems of Particles and Rotational Motion

Centre of Mass

A circular disc of radius R...

Physics

A circular disc of radius R is removed from a bigger circular disc of radius 2R such that the circumferences of the discs coincide. The centre of mass of the new disc is at a distance of αR from the centre of the bigger disc. The value of α is _______.

Medium

Answer

Let the mass per unit area of the disc be σ

radius is 2R

then mass of the original disc is M=π(2R)

2

σ=4πR

2

σ

Radius of the disc cut is R

them mass of the smaller disc M

=πR

2

σ=

4

M

Let O and O

be the centers of the original disc and the disc cut off from the original.

It is given that OO

=R

After the smaller disc has been cut from the original, the remaining portion is considered to be system of two masses. The two masses are M and −M

=−

4

M

The negative sign indicates the mass has been removed from the original disc.

Let x be the distance through which the center of mass of the remaining portion shifts from point O

the relation between the center of masses of two masses as

x=

m

1

+m

2

m

1

r

1

+m

2

r

2

x=

M−M

M×0−M

×R

x=

M−

4

M

M×0−

4

M

×R

x=

4

3M

4

MR

=

3

−R

therefore α=

3

1

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